Solved – What test – close to zero

Sounds as if you don't care whether an angle is positive or negative–only how far it is from zero. So you'd want to take the absolute value of each angle before conducting your test. And the natural candidates for that would be a T-test if you have large amounts of data (implying the sampling distributions of the mean absolute values would be approximately normal) and the Mann-Whitney U if you don't. (@Stephane's suggestion of ANOVA amounts to a T-test when you have only 2 groups.)

This R code illustrates the Mann-Whitney procedure.

# Create some data. set.seed(17) males <- rnorm(32) females <- rnorm(32) * 3/2  # The Wilcoxon/Mann-Whitney test on absolute values. wilcox.test(abs(males), abs(females)) 

The result in this case is a Wilcoxon statistic of 358 for two groups of 32 observations, giving a p-value of 0.0387: because it is less than a conventional threshold of 0.05, it can be taken as some evidence that the female deviations are greater than the male deviations. To get a better picture of these data, let's look at histograms (red=female, cyan=male):

maleHist <- hist(males, freq=FALSE) femaleHist <- hist(females, freq=FALSE) xMin <- min(males, females) xMax <- max(males, females) yMax <- max(maleHist$intensities, femaleHist$intensities) plot(femaleHist, freq=FALSE, xlim=c(xMin, xMax), ylim=c(0, yMax), col=hsv(1, alpha=0.5), main="Histograms", xlab="Angle (degrees)") lines(maleHist, freq=FALSE, col=hsv(.5, alpha=0.5)) 

Evidently, about 32 values in each group are needed to distinguish these sets of deviations, one of which is about 50% greater in size than the other: your power to tell that one group of deviations is closer to zero than the other is not very good.